$x^{3}-x^{2}-\alpha\left ( x-1 \right )$ bifurcation diagram

Question

I have a one-dim flow:

$\dot{x}=x^{3}-x^{2}-\alpha\left ( x-1 \right )$

I would like to attempt to sketch the flow and bifurcation diagram for this. Without the use of any mathematical software, are there tricks to dealing with such an equation?

Any help is appreciated.

Answer 1

Let $f(x) = x^3 − x^2 − \alpha (x−1)$. Factorising $f(x) = (x - 1)(x^2 - \alpha)$, we see that the fixed points are $x = 1$ (for all $\alpha$) and $\pm\sqrt{\alpha}$ (for $\alpha \ge 0$).

To determine their stability, we don't factorise. $f'(x) = 3x^2 - 2x - \alpha$, so that

$$ \begin{align} f'(1) &= 1 - \alpha \\ f'(\pm\sqrt{\alpha}) &= 2\alpha \mp 2\sqrt{\alpha} = 2\sqrt{\alpha} (\sqrt{\alpha} \mp 1) \end{align} $$

For flows, stability of fixed points is determined by the sign of $f'$. Hence

• $x = 1$ is unstable for $\alpha < 1$
• $x = +\sqrt{\alpha}$ is stable for $0 < \alpha < 1$
• $x = -\sqrt{\alpha}$ is unstable for all $\alpha > 0$

which is enough information to sketch a bifurcation diagram from.

EDIT. As a check on the the above, you can also (easily) sketch the graph of $f(x) = (x - 1)(x^2 - \alpha)$ for $\alpha < 0$, for $0 < \alpha < 1$ and for $\alpha > 1$. Alternatively, observe that for any cubic with positive leading coefficient and three distinct real roots, the slope is negative at the middle root and positive at the other two. So for this particular problem, whenever there are three distinct fixed points, only the middle one is stable; now just work out which one is the middle one.